Geometric subgroups of surface braid groups

Luis Paris and Dale Rolfsen

Abstract. Let M be a surface, let N be a subsurface of M , and let n ≤ m be

two positive integers. The inclusion of N in M gives rise to a homomorphism from the

braid group BnN with n strings on N to the braid group BmM with m strings on

M . We first determine necessary and sufficient conditions that this homomorphism is

injective, and we characterize the commensurator, the normalizer and the centralizer of

π1N in π1M . Then we calculate the commensurator, the normalizer, and the centralizer

of BnN in BmM for large surface braid groups.

1. Introduction

The classical braid groups Bm were introduced by Artin in 1926 ([Ar1], [Ar2]) andhave played a remarkable role in topology, algebra, analysis, and physics. A naturalgeneralization to braids on surfaces was introduced by Fox and Neuwirth [FoN] in 1962.The surface braid groups, for closed surfaces, were calculated in terms of generators andrelations during the ensuing decade ([Bi1], [Sc], [Va], [FaV]). Since then, most progressin this subject has been in its relation with mapping class groups and the general theoryof configuration spaces (see the surveys [Bi3], [Co]). However, recently there is renewedinterest in these fascinating groups in their own right, in part because of the action ofsurface braid groups on certain topological quantum field theories.

The purpose of this paper is to continue the study of the structure of the surfacebraid groups, with emphasis on certain naturally-occuring subgroups. A subsurface of asurface gives rise to inclusion maps between their braid groups. We determine necessaryand sufficient conditions that these inclusion-induced maps are injective in Section 2. Theremainder of the paper is devoted to a detailed study of these “geometric” subgroups. Inparticular, we calculate their centralizers, normalizers and commensurators in the largersurface braid group. Commensurators, in infinite groups, are of importance in their (uni-tary) representation theory. It is our hope that these results will be useful in the furtherstudy of surface braid groups, their representations and applications. In the remainder ofthis introductory section we present definitions, basic properties of surface braid groups,and a brief review of the literature.

1.1. Surface braids and configuration spaces

Let M be a topological manifold and choose distinct points P1, . . . , Pm ∈M (later wewill specialize to dim(M) = 2). A braid with m strings on M based at (P1, . . . , Pm) is anm-tuple b = (b1, . . . , bm) of paths, bi : [0, 1]→M , such that

Mathematics Subject Classification. Primary 20F36, 57M05. Secondary 57M60.

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1) bi(0) = Pi and bi(1) ∈ {P1, . . . , Pm} for all i ∈ {1, . . . , m},2) bi(t) 6= bj(t) for i, j ∈ {1, . . . , m}, i 6= j, and for t ∈ [0, 1].There is a natural notion of homotopy of braids. The braid group with m strings on

M based at (P1, . . . , Pm) is the group BmM = BmM(P1, . . . , Pm) of homotopy classes ofbraids based at (P1, . . . , Pm). The group operation is concatenation of braids, generalizingthe construction of the fundamental group. Indeed, for the case m = 1 we clearly haveB1M(P1) = π1(M,P1). For m > 1, it is useful to consider the class of pure braids, whichhave the property bi(1) = Pi. These form a subgroup of Bm which we will denote byPBmM = PBmM(P1, . . . , Pm). Let Σm be the group of permutations of {P1, . . . , Pm}.There is a natural epimorphism σ : BmM → Σm; its kernel is the pure braid group, so wehave an exact sequence:

1 −→ PBmM −→ BmMσ−→Σm −→ 1.

Note that, if M is a connected manifold of dimension at least two, then BmM andPBmM do not depend (up to isomorphism) on the choice of P1, . . . , Pm. An m-braidnaturally gives rise to m different paths in M under the map b→ (b1, . . . , bm). In the caseof pure braids these are loops, so there is a natural homomorphism

PBmM −→ π1(M,P1)× · · · × π1(M,Pm) ∼= π1(Mm),

where Mm denotes the m-fold cartesian power.

Proposition 1.1 ([Bi1]). If M is a connected manifold with dim(M) > 2, the above mapis an isomorphism. For dim(M) = 2 it is surjective.

The proof is straightforward when one views braids from the configuration space pointof view ([FoN], [FaN].) Let FmM denote the space of (ordered) distinct points of M , inother words FmM = (Mm) \ V , where V is the big diagonal, consisting of m-tuplesx = (x1, . . . , xm) for which xi = xj for some i 6= j. Then we clearly have an isomorphism

PBmM ∼= π1(FmM).

Proof of Proposition 1.1. The map in question is induced by the inclusion FmM =(M)m \ V → (M)m. Noting that V = ∪1≤i<j≤m{xi = xj} is a union of submanifoldsof codimension dim(M), the proposition follows from well-known general position argu-ments. ut

Because of Proposition 1.1, braid theory (as formulated here) is of marginal interestfor dimension ≥ 3 and we concentrate on dimension two, i. e. surface braid groups.

In the remainder of the paper, M will denote a connected surface, possibly withboundary and possibly nonorientable. To avoid pathology, we will assume M is eithercompact, or at least that it is a “punctured” compact manifold, i. e. M is homeomorphicto a compact 2-manifold, possibly with a finite set of points removed.

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By permuting coordinates, there is a natural action of Σm upon FmM and we denotethe orbit space, the space of unordered m-tuples, or configuration space, by FmM =FmM/Σm. We may view the full braid group as its fundamental group

BmM ∼= π1(FmM).

The inclusion PBmM ⊆ BmM may thus be interpreted as the mapping induced bythe covering space map FmM −→ FmM , which has fiber Σm. Fox and Neuwirth notedthat Bm(D), the braid groups of the disk D2, coincide with the Artin braid groups.

One of the most useful tools in studying braid groups is the Fadell-Neuwirth fibrationand its generalizations. As observed in [FaN], if M is a manifold and 1 ≤ n < m the mapρ : FmM → FnM defined by ρ(x1, . . . , xm) = (x1, . . . , xn) is a (locally trivial) fibrationwhich has the fiber Fm−n(M \ {P1, . . . , Pn}). This gives rise to a long exact sequence ofhomotopy groups of these spaces. For example, in the case n = m − 1 we have the exactsequence

· · · → π2FmM → π2Fm−1M → π1(M \ {P1, . . . , Pm−1})→ PBmM → PBm−1M → 1.

The punctured surface M\{P1, . . . , Pm−1} has the homotopy type of a one-dimensionalcomplex, and we see immediately from the above long exact sequence that

πk(FmM) ∼= πk(Fm−1M) ∼= · · · ∼= πk(M), k ≥ 3

andπ2(FmM) ⊂ π2(Fm−1M) ⊂ · · · ⊂ π2(M).

Because they are the only surfaces with nontrivial higher homotopy groups, the sphereS2 and the projective plane P 2 are exceptional cases in the general theory.

Proposition 1.2. Suppose that M is a connected surface, M 6= S2 or P 2, and k ≥ 2.Then πkFmM and πkFmM are trivial groups.

Proof. Since FmM → FmM is a covering map, it suffices to prove the proposition forFmM . But this follows from the observations made above, since πk(M) = 1 for k ≥ 2. ut

Combining this with the Fadell-Neuwirth fibration:

Proposition 1.3. Suppose that M is a connected surface, M 6= S2 or P 2, and 1 ≤ n < m.There is an exact sequence

1 −→ PBm−nM \ {P1, . . . , Pn} −→ PBmMρ−→PBnM −→ 1 . ut

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Let Σn be the group of permutations of {P1, . . . , Pn} and let Σm−n be the group ofpermutations of {Pn+1, . . . , Pm}. The Fadell-Neuwirth map gives rise to a (locally trivial)fibration

ρ : FmM/(Σn × Σm−n) −→ FnM/Σn = FnM

which has the fiber

(Fm−nM \ {P1, . . . , Pn})/Σm−n = Fm−nM \ {P1, . . . , Pn} .

So:

Proposition 1.4. Suppose that M is a connected surface, M 6= S2 or P 2, and 1 ≤ n < m.There is an exact sequence

1 −→ Bm−nM \ {P1, . . . , Pn} −→ σ−1(Σn × Σm−n) −→ BnM −→ 1 . ut

1.2. Torsion

Except for M = S2, P 2, the configuration space FmM is an Eilenberg-Maclane space,i. e. a classifying space for BmM . As is well-known, a group which has elements of finiteorder must have an infinite-dimensional classifying space (see, e. g. [Br, Cap VIII]). SinceFmM has dimension 2m, we can then conclude.

Proposition 1.5. If M is a connected surface, M 6= S2 or P 2, then its braid groupsBmM have no elements of finite order.

The braid groups of S2 and P 2 do have torsion (with the exception the trivial groupB1(S2)). We give a quick review of these, following [FaV] and [Va]. For S2 take all thebasepoints to lie in a disk D2 ⊆ S2 and let σ1, . . . , σm−1 be the standard braid generatorsof Bm(D2); σi exchanges Pi and Pi+1. They satisfy the famous braid relations

σiσj = σjσi, |i− j| ≥ 2σiσi+1σi = σi+1σiσi+1, 1 ≤ i ≤ m− 2

(∗)

The same σi can also be taken to be generators of Bm(S2) where they still satisfy theserelations. The word σ1σ2 · · ·σm−1σm−1 · · ·σ2σ1 may be interpreted as the (pure) braid inwhich P1 circles around P2, . . . , Pm, while those points stay fixed. This is clearly homotopicin S2 to the identity braid, so we have the additional relation

σ1σ2 · · ·σm−1σm−1 · · ·σ2σ1 = 1.

It is shown in [FaV] that this, together with (*) are defining relations for Bm(S2). Theelement τ = σ1σ2 · · ·σm−1 has order 2m in Bm(S2); it can be pictured as a simple braidwhich permutes the basepoints cyclically.

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For the projective plane, take σi as above corresponding to a disk engulfing the base-points, and let ρj to be a braid in which the basepoint Pj travels along a nontrivial loop inP 2 while the other basepoints sit still. See [Va] for a more precise description and a proofthat Bn(P 2) is presented by the 2m− 1 generators σ1, . . . , σm−1, ρ1, . . . , ρm and relations(*) together with

σiρj = ρjσi, j 6= i, i+ 1ρi = σiρi+1σi

σ2i = ρ−1

i+1ρ−1i ρi+1ρi

ρ21 = σ1σ2 · · ·σm−1σm−1 · · ·σ2σ1.

The element τ as defined above, but considered an element of Bm(P 2), again has order2m. Thus we have the theorem of Van Buskirk, that for each m ≥ 2, the surface braidgroup BmM has elements of finite order if and only if M = S2 or P 2.

Some of these braid groups are actually finite: B2(S2) ∼= Z/2Z, B3(S2) has order 12,B1(P 2) ∼= Z/2Z and B2(P 2) is a group of order 16 whose subgroup PB2(P 2) is isomorphicwith the quaternion group {±1,±i,±j,±k}. B3(P 2) is infinite, as are all the other higherbraid groups of P 2 and S2.

For the braid groups of higher genus closed surfaces, we refer the reader to [Sc], andonly mention how to produce a generating set. After removing a disk from a surface Mof genus g, the remainder can be modelled as a disk with g twisted bands attached, in thenonorientable case, or 2g bands if the surface is orientable. Then Bm(M) is generated byσ1, . . . , σm−1 as above, plus ρij which represents the basepoint Pi running once around thejth band, while the others are fixed. A finite set of relations can be found in [Sc].

1.3. Centers and Large Surfaces

The center Z(G) of a group G is the subgroup of elements which commute with allelements of the group. Chow [Ch] proved that the groups Bm = Bm(D2) have infinitecyclic center, for m ≥ 2. Some other surface braid groups also have nontrivial centers:those of S2 [GV], P 2 [Va]. If τ is defined as in the preceding section, the element τm iscentral in BmS

2. Birman stated in [Bi2] that the torus braid groups BmT 2 have centerwhich is free abelian with two generators, but did not include a complete proof. We willprove this, and also calculate the center of the braid groups of the annulus S1×I in Section4. However, apart from these and a few other exceptions, most surface braid groups haveno center. Our proof is the same as given in [Bi2].

Definition. A compact surface M will be called large if M 6= S2, P 2, D2, S1 × I, T 2 =S1 × S1, Mobius strip S1×I, or Klein bottle S1×S1. In other words, we call a surfacelarge if its fundamental group has no finite index abelian subgroup.

Proposition 1.6. Let M be a large compact surface. Then the center Z(Bm(M)) is atrivial group.

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Proof. First, we prove by induction on m that Z(PBmM) = {1}. The case m = 1is well-known: the only surfaces whose fundamental groups have nontrivial centers areP 2, S1 × I, T 2, the Mobius strip, and Klein bottle.

Let m > 1 and M large. We consider the following exact sequence

1 −→ π1(M \ {P1, . . . , Pm−1}) −→ PBmMρ−→PBm−1M −→ 1.

Since ρ is surjective, it takes center into center, and by induction, Z(PBm−1M) = {1}.So Z(PBmM) ⊆ π1(M \ {P1, . . . , Pm−1}). But this latter group has trivial center, soZ(PBmM) = {1}. Now, let g ∈ Z(BmM). There exists an integer k > 0 such thatgk ∈ PBmM . Then gk ∈ Z(PBmM), thus gk = 1. By Proposition 1.5, g = 1. ut

2. Subsurfaces

A subsurface N of a surface M is the closure of an open subset of M . For simplicitywe make the extra assumption that every boundary component of N either is a boundarycomponent of M or lies in the interior of M .

Let P1 ∈ N . The inclusion N ⊆ M induces a morphism ψ : π1(N,P1) → π1(M,P1).The following proposition is well-known.

Proposition 2.1. Let N be a connected subsurface of M such that π1(N,P1) 6= {1}.The morphism ψ : π1(N,P1)→ π1(M,P1) is injective if and only if none of the connectedcomponents of the closure M \N of M \N is a disk. ut

Let P1, . . . , Pn ∈ N , and let Pn+1, . . . , Pm ∈M \N . The inclusion N ⊆M induces amorphism ψ : BnN → BmM .

Proposition 2.2. Let M be different from the sphere and from the projective plane,and let N be such that none of the connected components of M \N is a disk. Then themorphism ψ : BnN → BmM is injective.

Remark. Proposition 2.2 is proved in [Go] in the particular case where N is a disk.

Proof. Let Pψn : PBnN → PBnM be the morphism induced by the inclusion N ⊆ M .We prove that Pψn is injective by induction on n. The case n = 1 is a consequence ofProposition 2.1.

Let n > 1. By Proposition 2.1, the inclusion

N \ {P1, . . . , Pn−1} ⊆M \ {P1, . . . , Pn−1}

induces a monomorphism

α : π1(N \ {P1, . . . , Pn−1}) −→ π1(M \ {P1, . . . , Pn−1}) .

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The following diagram commutes.

1 −→ π1(N \ {P1, . . . , Pn−1}) −→ PBnNρ−→ PBn−1N −→ 1yα yPψn yPψn−1

1 −→ π1(M \ {P1, . . . , Pn−1}) −→ PBnMρ−→ PBn−1M −→ 1

By induction, Pψn−1 is injective. By the five lemma, Pψn is injective, too.Let Pψ : PBnN → PBmM be the morphism induced by the inclusion N ⊆ M . The

following diagram commutes.

PBnNPψ−→ PBmMyid

yρPBnN

Pψn−→ PBnM

The morphism Pψn is injective, thus Pψ is injective, too.Let ι : Σn → Σm be the inclusion. The following diagram commutes.

1 −→ PBnN −→ BnNσ−→ Σn −→ 1yPψ yψ yι

1 −→ PBmM −→ BmMσ−→ Σm −→ 1

Both Pψ and ι are injective, so, by the five lemma, ψ is injective, too. utLet N1, . . . , Nr be the connected components of M \N . For i = 1, . . . , r, we write

Pi = {Pn+1, . . . , Pm} ∩Ni.

Theorem 2.3. Let M be different from the sphere and from the projective plane. Themorphism ψ : BnN → BmM is injective if and only if either Ni is not a disk or Pi 6= ∅,for all i = 1, . . . , r.

Proof. We suppose that there exists i ∈ {1, . . . , r} such that Ni is a disk and such thatPi = ∅. We consider the following commutative diagram.

π1(N \ {P2, . . . , Pn}) −→ BnNyψ yψπ1(M \ {P2, . . . , Pm}) −→ BmM

By [FaN], the morphism π1(N \{P2, . . . , Pn})→ BnN is injective. On the other hand, themorphism ψ : π1(N \ {P2, . . . , Pn})→ π1(M \ {P2, . . . , Pm}) is clearly not injective. Thusψ : BnN → BmM is not injective.

We suppose that either Ni is not a disk or Pi 6= ∅, for all i = 1, . . . , r. We considerthe following commutative diagram.

BnNψ−→ BnM \ {Pn+1, . . . , Pm}yid

yBnN

ψ−→ BmM

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By Proposition 2.2, the morphism ψ : BnN → BnM \ {Pn+1, . . . , Pm} is injective. By[FaN], the morphism BnM \{Pn+1, . . . , Pm} → BmM is injective. Thus ψ : BnN → BmMis injective. ut

3. Commensurator, normalizer,

and centralizer of π1N in π1M

Let N be a subsurface of a connected surface M . We say that N is a Mobius collar inM if N is a cylinder S1× I and M \N has two components N1, N2 with one of them, sayN1, a Mobius strip (see Figure 3.1). Then M0 = N ∪ N1 will be called the Mobius stripcollared by N in M .

2

N

N

N

1

Figure 3.1

Let G be a group, and let H be a subgroup of G. We denote by CG(H) the commen-surator of H in G, by NG(H) the normalizer of H in G, and by ZG(H) the centralizer ofH in G. That is,

ZG(H) = {g ∈ G : gh = hg for all h ∈ H}

NG(H) = {g ∈ G : gHg−1 = H}

CG(H) = {g ∈ G : gHg−1 ∩H has finite index in gHg−1 and H}

The goal of this section is to prove the following theorem.

Theorem 3.1. Let P0 ∈ N . We write π1M = π1(M,P0) and π1N = π1(N,P0).i) If M is not large or if π1N = {1}, then Cπ1M (π1N) = π1M .ii) If M is large, if π1N 6= {1}, and if N is not a Mobius collar in M , then

Cπ1M (π1N) = π1N .iii) If M is large and if N is a Mobius collar in M , then Cπ1M (π1N) = π1M0, where

M0 is the Mobius strip collared by N in M .

Corollary 3.2. i) If M is either a cylinder, or a torus, or a Mobius strip, then

Cπ1M (π1N) = Nπ1M (π1N) = Zπ1M (π1N) = π1M .

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ii) If M is large, if N is not a Mobius collar in M , if π1N 6= {1}, and if N is notlarge, then

Cπ1M (π1N) = Nπ1M (π1N) = Zπ1M (π1N) = Z(π1N) = π1N .

iii) If M and N are both large, then

Cπ1M (π1N) = Nπ1M (π1N) = π1N ,

Zπ1M (π1N) = Z(π1N) = {1} .

iv) If M is large and if N is a Mobius collar in M , then

Cπ1M (π1N) = Nπ1M (π1N) = Zπ1M (π1N) = π1M0 ,

where M0 is the Mobius strip collared by N in M .

Before proving Theorem 3.1, we recall some well-known results on graphs of groups.An (oriented) graph Γ is the following data.1) A set V (Γ) of vertices.2) A set A(Γ) of arrows.3) A map s : A(Γ)→ V (Γ) called origin, and a map t : A(Γ)→ V (Γ) called end.A graph of groups G(Γ) on Γ is the following data.1) A group Gv for all v ∈ V (Γ).2) A group Ga for all a ∈ A(Γ).3) Two monomorphisms φa,s : Ga → Gs(a) and φa,t : Ga → Gt(a) for all a ∈ A(Γ).

We refer to [Se] for a general exposition on graphs of groups.Let T be a maximal tree of Γ. The fundamental group π1(G(Γ), T ) of G(Γ) based

at T is the (abstract) group given by the following presentation. The generating set ofπ1(G(Γ), T ) is

{ea ; a ∈ A(Γ)} ∪(∪v∈V (Γ)Gv

),

where {ea; a ∈ A(Γ)} is an abstract set in one-to-one correspondance with A(Γ). Therelations of π1(G(Γ), T ) are

1) the relations of Gv for all v ∈ V (Γ),2) ea = 1 for all a ∈ A(T ),3) e−1

a · φa,s(g) · ea = φa,t(g) for all a ∈ A(Γ) and for all g ∈ Ga.There is a morphism φv : Gv → π1(G(Γ), T ) for all v ∈ V (Γ). By [Se], this morphism isinjective.

The fundamental group π1(Γ, T ) of Γ based at T has the following presentation. Thegenerating set of π1(Γ, T ) is {ea; a ∈ A(Γ)}. The set of relations of π1(Γ, T ) is {ea = 1; a ∈A(T )}.

Let p : Γ → Γ be the universal cover of Γ. Let G(Γ) be the graph of groups on Γdefined as follows.

1) Gv = Gp(v) for all v ∈ V (Γ).2) Ga = Gp(a) for all a ∈ A(Γ).

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3) φa,s = φp(a),s and φa,t = φp(a),t for all a ∈ A(Γ).We fix a section S : T → Γ of p over T . We extend S to a section S : A(Γ) → A(Γ)

as follows. Let a ∈ A(Γ). Then S(a) is the unique lift of a such that t(S(a)) = S(t(a)).We define an action of π1(Γ, T ) on π1(G(Γ), Γ) as follows. Let v ∈ V (Γ), let g ∈ Gv,

and let u ∈ π1(Γ, T ). Thenu(g) = g ∈ Gu(v) .

We consider the corresponding semidirect product π1(G(Γ), Γ)×7 π1(Γ, T ). By [Se], thereis an isomorphism

π1(G(Γ), T ) −→ π1(G(Γ), Γ)×7 π1(Γ, T )

which sends Gv isomorphically on GS(v) for all v ∈ V (Γ), and which sends ea on ea for alla ∈ A(Γ). So, we can assume that π1(G(Γ), T ) = π1(G(Γ), Γ)×7 π1(Γ, T ), that Gv = GS(v)

for all v ∈ V (Γ), and that Ga = GS(a) for all a ∈ A(Γ).Let G = π1(G(Γ), T ), and let G = π1(G(Γ), Γ). The universal cover of G(Γ) is the

graph Γ defined as follows.V (Γ) = (V (Γ)× G)/ ∼ ,

where ∼ is the equivalence relation defined by

(v1, g1) ∼ (v2, g2) if v1 = v2 = v and g−12 g1 ∈ Gv .

We denote by [v, g] the equivalence class of (v, g).

A(Γ) = (A(Γ)× G)/ ∼ ,

where ∼ is the equivalence relation defined by

(a1, g1) ∼ (a2, g2) if a1 = a2 = a and g−12 g1 ∈ Ga .

We denote by [a, g] the equivalence class of (a, g). The origin map s : A(Γ) → V (Γ) isdefined by

s([a, g]) = [s(a), g]

for a ∈ A(Γ) and for g ∈ G. The end map t : A(Γ)→ V (Γ) is defined by

t([a, g]) = [t(a), g]

for a ∈ A(Γ) and for g ∈ G. By [Se], Γ is a tree.The group G acts on Γ as follows. Let u ∈ π1(Γ, T ), let h, g ∈ G, let v ∈ V (Γ), and

let a ∈ A(Γ). Then

h([v, g]) = [v, hg] , h([a, g]) = [a, hg] ,u([v, g]) = [u(v), ugu−1] , u([a, g]) = [u(a), ugu−1] .

The isotropy subgroup of a vertex v ∈ V (Γ) is

Isot(v) = {g ∈ G ; g(v) = v} .

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The isotropy subgroup of an arrow a ∈ A(Γ) is

Isot(a) = {g ∈ G ; g(a) = a} .

Let v ∈ V (Γ) and let a ∈ A(Γ). By [Se],

Isot([S(v), 1]) = Gv ,

Isot([S(a), 1]) = Ga .

Now, we come back to our original assumptions. M is a surface (with boundary)different from the sphere and from the projective plane. N is a subsurface of M such thatnone of the connected components of M \N is a disk. Without lost of generality, we canalso assume that N is not a disk. Let N1, . . . , Nr be the connected components of M \N .

We define a graph Γ as follows.

V (Γ) = {v0, v1, . . . , vr} .

For i ∈ {1, . . . , r}, we fix an abstract set Ai(Γ) in one-to-one correspondance with theconnected components of N ∩Ni. We set

A(Γ) = ∪ri=1Ai(Γ) .

If a ∈ Ai(Γ), then s(a) = v0 and t(a) = vi.We define a graph of groups G(Γ) on Γ as follows. Let i ∈ {1, . . . , r}. We fix a point

Pi ∈ Ni and we setGvi = Gi = π1(Ni, Pi) .

We fix a point P0 ∈ N and we set

Gv0 = G0 = π1(N,P0) .

Let a ∈ Ai(Γ). We denote by Ca the connected component of N ∩Ni which correspondsto a. The set Ca is a boundary component of both N and Ni. We fix a point Pa ∈ Ca andwe set

Ga = π1(Ca, Pa) ' Z .

We fix a path γa,s : [0, 1] → N from P0 to Pa. This path induces a monomorphismφa,s : Ga → G0. We fix a path γa,t : [0, 1] → Ni from Pi to Pa. This path induces amonomorphism φa,t : Ga → Gi.

We fix an arrow ai ∈ Ai(Γ) for all i ∈ {1, . . . , r}. We consider the graph T defined asfollows.

1) V (T ) = {v0, v1, . . . , vr}.2) A(T ) = {a1, . . . , ar}.3) s(ai) = v0 and t(ai) = vi for all i ∈ {1, . . . , r}.

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The graph T is a maximal tree of Γ. We write

γa = γa,sγ−1a,t

βa = γaγ−1ai

for all a ∈ Ai(Γ). For i = 1, . . . , r, the path γai induces a morphism

ψi : Gi = π1(Ni, Pi) −→ π1(M,P0) .

We denote byψ0 : G0 = π1(N,P0) −→ π1(M,P0)

the morphism induced by the inclusion N ⊆M .The following theorem is a well-known version of Van Kampen’s theorem.

Theorem 3.3. The map

{ea; a ∈ A(Γ)} −→ π1(M,P0)ea 7−→ βa

and the morphisms ψi : Gi → π1(M,P0) (i = 0, 1, . . . , r) induce an isomorphism

ψ : π1(G(Γ), T ) −→ π1(M,P0) .

Let Γ be the universal cover of G(Γ). Let q : Γ → Γ be the map defined as follows.Let v ∈ V (Γ), let a ∈ A(Γ), and let g ∈ G. Then

q([v, g]) = p(v) ,q([a, g]) = p(a) .

The following lemma is a preliminary result to the proof of Theorem 3.1.

Lemma 3.4. Let i ∈ {1, . . . , r}. Let v ∈ V (Γ) be such that q(v) = vi. Let a, b ∈ A(Γ) besuch that t(a) = t(b) = v (see Figure 3.2).

i) If q(a) = q(b) and Isot(a) ∩ Isot(b) 6= {1}, then Ni is a Mobius strip.ii) If q(a) 6= q(b) and Isot(a)∩ Isot(b) 6= {1}, then Ni is a cylinder and both boundary

components of Ni are included in N ∩Ni.

����

��������

����

����

����

a_

_b_

v

Figure 3.2

12

Proof. i) We suppose that i = 1 and that a = [S(a1), 1]. Then

v = t(a) = [t(S(a1)), 1] = [S(v1), 1] .

Let b = [b, g]. Thent(b) = [t(b), g] = [S(v1), 1] ,

thus b = S(a1) (since q(a) = q(b) = a1), and g ∈ Gv1 = G1. Note that g 6∈ Ga1 , otherwise

b = [S(a1), g] = [S(a1), 1] = a .

So,Isot(a) = Ga1 and Isot(b) = gGa1 g

−1 .

Let h1 be a generator of Ga1 . There exist k1, k2 ∈ Z \ {0} such that

hk11 = ghk2

1 g−1 .

We suppose that N1 is not a Mobius strip. Let F be the subgroup of G1 generated byh1 and g. The subsurface N1 has non-empty boundary, thus G1 is a free group, thereforeF is a free group of rank either 1 or 2. Since F is a hopfian group (see [LS, Prop. 3.4])and since hk1

1 = ghk21 g−1, the group F has rank 1. By [Ep, Thm. 4.2], h1 generates F . In

particular, there exists l ∈ Z such that

g = hl1 ∈ Ga1 .

This is a contradiction. So, N1 is a Mobius strip.

ii) We suppose that i = 1 and that a = [S(a1), 1]. Then

v = t(a) = [t(S(a1)), 1] = [S(v1), 1] .

Let b = [b, g] and let b = q(b) 6= a1. Then

t(b) = [t(b), g] = [S(v1), 1] ,

thus b = S(b) and g ∈ Gv1 = G1. So,

Isot(a) = Ga1 and Isot(b) = gGbg−1 .

Let h1 be a generator of Ga1 , and let h be a generator of Gb. There exist k1, k2 ∈ Z \ {0}such that

hk11 = ghk2 g−1 .

Let F be the subgroup of G1 generated by h1 and ghg−1. Since G1 is a free group, F isa free group of rank either 1 or 2. Since F is a hopfian group and since hk1

1 = (ghg−1)k2 ,the group F has rank 1. The subsurface N1 has at least two boundary components, Ca1

13

and Cb, thus N1 is not a Mobius strip. By [Ep, Thm. 4.2], both h1 and ghg−1 generateF . So, we can assume that

h1 = ghg−1 .

By [Ep, Lemma 2.4], it follows that N1 is a cylinder and that Ca1 and Cb are the boundarycomponents of N1. ut

Proof of Theorem 3.1. i) It is obvious, as all the non-large surfaces have abelian funda-mental groups, except the Klein bottle, which has an abelian subgroup of index 2.

ii) We suppose that there exists g ∈ Cπ1M (π1N) such that g 6∈ π1N , and we provethat either M is not large, or N is a Mobius collar in M .

Let v0 = [S(v0), 1] ∈ V (Γ). We have g(v0) 6= v0 since g 6∈ π1N = Isot(v0). Let

aε11 aε22 . . . aεll (ai ∈ A(Γ) and εi ∈ {±1})

be the (unique) reduced path of Γ from v0 to g(v0) (see Figure 3.3). For j = 1, . . . , l wedenote by vj the end of the path aε11 . . . a

εjj . Note that l ≥ 2 since q(g(v0)) = q(v0) = v0. If

h ∈ G0∩gG0g−1, then h ∈ Isot(v0) and h ∈ Isot(g(v0)), thus h ∈ Isot(vj) and h ∈ Isot(aj)

for all j ∈ {1, . . . , l}. We suppose that q(v1) = v1.

{1} 6= G0 ∩ gG0g−1 ⊆ Isot(a1) ∩ Isot(a2) ,

thus, by Lemma 3.4, either N1 is a Mobius strip, or N1 is a cylinder and both boundarycomponents of N1 are included in N ∩N1.

������

������

������

������

������

������

������

������

����

����

����

a a a_ _ _

v������

������

v_

v v_ _ _ _

1 2 l

0 1 2 l-1 g(v 0)

Figure 3.3

The group G0 ∩ gG0g−1 has finite index in G0 = π1N , it is included in Isot(a1), and

Isot(a1) is an infinite cyclic group. So, π1N has an infinite cyclic subgroup of finite index,thus either N is a cylinder, or N is a Mobius strip.

If N is a Mobius strip, then N1 is also a Mobius strip and M = N ∪ N1 is a Kleinbottle (see Figure 3.4).

1N N

Figure 3.4

14

If N and N1 are both cylinders, then M = N ∪N1 is a torus (see Figure 3.5).

1N N

Figure 3.5

If N is a cylinder and if N1 is a Mobius strip, then N is a Mobius collar in M (seeFigure 3.6).

2

N

N

N

1

Figure 3.6

iii) We suppose that N is a cylinder, that N1 is a Mobius strip, and that M is large(see Figure 3.6). Let M0 = N∪N1 be the Mobius strip collared by N in M . The subsurfaceM0 is not a Mobius collar in M , thus, by ii),

Cπ1M (π1M0) = π1M0 .

The group π1N has finite index in π1M0, thus

Cπ1M (π1N) = Cπ1M (π1M0) = π1M0 . ut

4. centers

The goal of this section is to describe the center of BmM , where M is either a cylinderor a torus.

15

Let C be a cylinder. We assume that

C = {z ∈ C ; 1 ≤ |z| ≤ 2} ,

and thatPi = 1 +

i

m+ 1for i = 1, . . . , m .

Let di : [0, 1]→ C be the path defined by

di(t) =(

1 +i

m+ 1

)e2iπt for t ∈ [0, 1] .

Let α be the element of PBmC represented by d = (d1, . . . , dm) (see Figure 4.1).

p p p.

d

d

d

1 2 m

1

2

m

. .

Figure 4.1

Proposition 4.1. With the above assumptions, the center of BmC is the infinite cyclicsubgroup generated by α.

Proof. LetD = {z ∈ C ; |z| ≤ 2} .

Let P0 = 0. The inclusion C ⊆ D \ {P0} induces an isomorphism BmC → Bm(D \ {P0}).Let Σm+1 be the group of permutations of {P0, P1, . . . , Pm}, and let Σm be the group

16

of permutations of {P1, . . . , Pm}. We consider the morphism σ : Bm+1D → Σm+1. ByProposition 1.4, we have the following exact sequence.

1 −→ Bm(D \ {P0}) −→ σ−1(Σm)−→π1(D,P0) −→ 1

Moreover, π1(D,P0) = {1}. Thus the inclusion D \ {P0} ⊆ D induces an isomorphismBm(D \ {P0}) → σ−1(Σm). The image of α by this isomorphism is the element ofBm+1D, denoted by α, represented by the braid b = (P0, d1, . . . , dm). By [Ch], we haveZ(Bm+1D) = Z(PBm+1D), and this group is the infinite cyclic subgroup generated by α.From the inclusions

PBm+1D ⊆ σ−1(Σm) ⊆ Bm+1D ,

it follows that the center of σ−1(Σm) is equal to the center of Bm+1D which is the cyclicsubgroup generated by α. Thus, by the preceding isomorphism, the center of BmC =Bm(D \ {P0}) is the infinite cyclic subgroup generated by α. ut

Now, we describe the center of BmT , where T is a torus. We assume that

T = R2/Z2 .

We denote by (x, y) the equivalence class of (x, y). We assume that

Pi =(i+ 1m+ 3

,i+ 1m+ 3

)for i = 1, . . . , m .

Let ai : [0, 1]→ T be the path defined by

ai(t) =(i+ 1m+ 3

− t, i+ 1m+ 3

)for t ∈ [0, 1] ,

and let bi : [0, 1]→ T be the path defined by

bi(t) =(i+ 1m+ 3

,i+ 1m+ 3

− t)

for t ∈ [0, 1] .

Let α be the element of PBmT represented by a = (a1, . . . , am) (see Figure 4.2), and letβ be the element of of PBmT represented by b = (b1, . . . , bm).

17

P

P

P.

a

a

a

1

2

m

1

2

m.

.

Figure 4.2

Proposition 4.2. With the above assumptions, the center of BmT is the subgroup gen-erated by α and β. It is a free abelian group of rank 2.

Proof. The proof of Proposition 4.2 is divided into 4 steps. Let Zm denote the subgroupof PBmT generated by α and β.

Step 1. Zm is a free abelian group of rank 2.

By [Bi1, Thm. 5], α and β commute, thus Zm is an abelian group. We consider thefollowing exact sequence.

1 −→ PBm−1T \ {P1} −→ PBmTρ−→π1(T, P1) −→ 1

The group π1(T, P1) is a free abelian group of rank 2 and {ρ(α), ρ(β)} is a basis of π1(T, P1),thus Zm is also a free abelian group of rank 2.

Step 2. Zm ⊆ Z(BmT ).

Let

D =[

1m+ 3

,m+ 2m+ 3

]×[

1m+ 3

,m+ 2m+ 3

]⊆ T .

By Proposition 2.2, the inclusion D ⊆ T induces a monomorphism BmD → BmT . Thefollowing diagram commutes.

1 −→ PBmD −→ BmDσ−→ Σm −→ 1y y yid

1 −→ PBmT −→ BmTσ−→ Σm −→ 1

Thus BmT is generated by PBmT ∪BmD.By [Bi1, Thm. 5], α commutes with all the elements of PBmT .Let

C =(

R×[

1m+ 3

,m+ 2m+ 3

])/Z ⊆ T .

18

By Proposition 2.2, the inclusion C ⊆ T induces a monomorphism BmC → BmT . More-over, α ∈ BmC and BmD ⊆ BmC. By Proposition 4.1, Z(BmC) is the infinite cyclicsubgroup generated by α. So, α commutes with all the elements of BmD.

This shows that α ∈ Z(BmT ). Similarly, β ∈ Z(BmT ).

Step 3. Z(PBmT ) ⊆ Zm.

We prove Step 3 by induction on m. Let m = 1. Then PB1T = π1(T, P1) = Z1, thusZ(PB1T ) = Z1.

Let m > 1. Let g ∈ Z(PBmT ). We consider the following exact sequence.

1 −→ π1(T \ {P1, . . . , Pm−1}) −→ PBmTρ−→PBm−1T −→ 1

We have ρ(g) ∈ Z(PBm−1T ). By induction, Z(PBm−1T ) ⊆ Zm−1. Moreover, ρ(Zm) =Zm−1. Thus we can choose h ∈ Zm such that ρ(h) = ρ(g). We write g′ = gh−1.Then g′ ∈ Z(PBmT ) and g′ ∈ π1(T \ {P1, . . . , Pm−1}) (since ρ(g′) = 1), thus g′ ∈Z(π1(T \ {P1, . . . , Pm−1})) = {1}, thus g′ = gh−1 = 1, therefore g = h ∈ Zm.

Step 4. Z(BmT ) ⊆ PBmT .

Let g ∈ BmT . We suppose that there exist i, j ∈ {1, . . . , m}, i 6= j such thatσ(g)(Pi) = Pj, and we prove that g 6∈ Z(BmT ).

Let αi ∈ PBmT represented by (P1, . . . , Pi−1, ai, Pi+1, . . . , Pm), where Pk denotes theconstant path on Pk for k = 1, . . . , m and ai is as above. We consider the following exactsequence.

1 −→ PBm−1T \ {Pi} −→ PBmTρi−→π1(T, Pi) −→ 1

Then ρi(αi) 6= 1 and ρi(gαig−1) = 1 (see Figure 4.3), thus gαig−1 6= αi, therefore g 6∈Z(BmT ). ut

-1

..

...

..

..

...

..

..

...

..

..

...

..

P

PP

Pi

j

i

j

g gα i

Figure 4.3

19

5. Commensurator, normalizer, and

centralizer of BnD in BmM

Let M be an oriented surface different from the sphere, and let D ⊆ M be a diskembedded in M . Let n ≥ 2, let P1, . . . , Pn ∈ D, and let Pn+1, . . . , Pm ∈ M \ D. Thegoal of this section is to describe the commensurator, the normalizer, and the centralizerof BnD in BmM . Note that, if n = 1, then B1D = {1}, thus

CBmM (B1D) = NBmM (B1D) = ZBmM (B1D) = BmM .

This section is divided into two subsections. We state our results in Subsection 5.1,and we prove them in Subsection 5.2.

5.1. Statements

A tunnel on M based at (D;Pn+1, . . . , Pm) is a map

H : D ∪ {Pn+1, . . . , Pm} × [0, 1] −→M

such that1) H(x, 0) = H(x, 1) = x for all x ∈ D,2) H(Pi, 0) = Pi and H(Pi, 1) ∈ {Pn+1, . . . , Pm} for all Pi ∈ {Pn+1, . . . , Pm},3) H(x, t) 6= H(y, t) for x, y ∈ D ∪ {Pn+1, . . . , Pm}, x 6= y, and for t ∈ [0, 1].There is a natural notion of homotopy of tunnels. The tunnel group on M based at

(D;Pn+1, . . . , Pm) is the group Tm−nM = Tm−nM(D;Pn+1, . . . , Pm) of homotopy classesof tunnels on M based at (D;Pn+1, . . . , Pm). Multiplication is concatenation, as withbraids.

We define a morphism

τ : Tm−nM ×BnD −→ BmM

as follows. Let h ∈ Tm−nM and let f ∈ BnD. Let H be a tunnel on M based at(D;Pn+1, . . . , Pm) which represents h, and let b = (b1, . . . , bn) be a braid on D based at(P1, . . . , Pn) which represents f . Let b = (b1, . . . , bn, bn+1, . . . , bm) be the braid on Mdefined by

bi(t) = H(bi(t), t) for i ∈ {1, . . . , n} and for t ∈ [0, 1],bi(t) = H(Pi, t) for i ∈ {n+ 1, . . . , m} and for t ∈ [0, 1].

Then τ(h, f) is the element of BmM represented by b.

Remark. This is related to the tensor product operation for the classical braid groups (see[Co]).

We denote by Cn,mM the image of τ . Let h ∈ Tm−nM and let f, f ′ ∈ BnD. Then

τ(h, f) · f ′ · τ(h, f)−1 = τ(1, ff ′f−1) = ff ′f−1 .

20

In particular,Cn,mM ⊆ NBmM (BnD) .

Theorem 5.1. Let n ≥ 2, and M be an orientable surface, M 6= S2. Then

CBmM (BnD) = Cn,mM .

Let Zn,mM denote the image by τ of Tm−nM × Z(BnD).

Corollary 5.2. Let n ≥ 2. Then

CBmM (BnD) = NBmM (BnD) = Cn,mM ,

ZBmM (BnD) = Zn,mM .

Remark. i) We do not know whether a similar result holds for non-orientable surfaces.ii) Corollary 5.2 generalizes [FRZ, Thm. 4.2].

Let B1m−n+1M = B1

m−n+1M(P1;Pn+1, . . . , Pm) denote the subgroup of Bm−n+1M =Bm−n+1M(P1, Pn+1, . . . , Pm) consisting of g ∈ Bm−n+1M such that σ(g)(P1) = P1. Wedefine a morphism κ : Tm−nM → B1

m−n+1M as follows. Let h ∈ Tm−nM . Let H be atunnel on M based at (D;Pn+1, . . . , Pm) which represents h. Let b = (b1, bn+1, . . . , bm) bethe braid defined by

bi(t) = H(Pi, t) for i ∈ {1, n+ 1, . . . , m} and for t ∈ [0, 1].Then κ(h) is the element of B1

m−n+1M represented by b.

Theorem 5.3. Let n ≥ 2. There exists a morphism δ : Cn,mM → B1m−n+1M such that

δ(τ(h, f)) = κ(h)

for all h ∈ Tm−nM and for all f ∈ BnD. Moreover, we have the following exact sequences.

1 −→ BnD −→ Cn,mMδ−→B1

m−n+1M −→ 11 −→ Z(BnD) −→ Zn,mM

δ−→B1m−n+1M −→ 1

Theorem 5.4. Let n ≥ 2. Let M be either with non-empty boundary or a torus. Thereexists a morphism ι : B1

m−n+1M → Zn,mM such that δ ◦ ι = id. In particular,

Cn,mM ' B1m−n+1M ×BnD ,

Zn,mM ' B1m−n+1M × Z(BnD) .

21

Remark. Theorem 5.4 generalizes [FRZ, Thm. 4.3] and [Ro, Thm. 3].

5.2. Proofs

Lemma 5.5. We consider an exact sequence

1 −→ G1 −→ G2φ−→G3 −→ 1

Let H2 ⊆ G2 be a subgroup, let H3 = φ(H2), and let H1 = H2 ∩G1. Then

φ(CG2(H2)) ⊆ CG3(H3) ,CG2(H2) ∩G1 ⊆ CG1(H1) .

Proof. Let g ∈ CG2(H2). We write

F2 = H2 ∩ gH2g−1 .

Let h1, . . . , hk ∈ H2 be such that

H2 = F2 ∪ h1F2 ∪ . . . ∪ hkF2 .

Thenφ(H2) = H3 = φ(F2) ∪ φ(h1)φ(F2) ∪ . . . ∪ φ(hk)φ(F2) .

So, φ(F2) has finite index in H3. Moreover,

φ(F2) = φ(H2 ∩ gH2g−1) ⊆ φ(H2) ∩ φ(gH2g

−1) = H3 ∩ φ(g)H3φ(g)−1 ,

thus H3 ∩ φ(g)H3φ(g)−1 has finite index in H3. Similarly, H3 ∩ φ(g)H3φ(g)−1 has finiteindex in φ(g)H3φ(g)−1. So, φ(g) ∈ CG3(H3).

Let g ∈ CG2(H2) ∩G1. We write

F2 = H2 ∩ gH2g−1 .

Let h1, . . . , hk ∈ H2 be such that

H2 = F2 ∪ h1F2 ∪ . . . ∪ hkF2 .

We assume that

hiF2 ∩H1 6= ∅ for i = 1, . . . , l ,hiF2 ∩H1 = ∅ for i = l + 1, . . . , k .

22

We can also assume that hi ∈ H1 for i = 1, . . . , l. Then

H1 = (F2 ∩H1) ∪ h1(F2 ∩H1) ∪ . . . ∪ hl(F2 ∩H1) .

Moreover,F2 ∩H1 = H2 ∩ gH2g

−1 ∩H1 = H1 ∩ gH1g−1 .

Thus H1 ∩ gH1g−1 has finite index in H1. Similarly, H1 ∩ gH1g

−1 has finite index ingH1g

−1. So, g ∈ CG1(H1). ut

Lemma 5.6. Let M be either with non-empty boundary or a torus. There exists a morphismι0 : B1

m−n+1M → Tm−nM such that κ ◦ ι0 = id.

Proof. Let TM be the tangent space of M . It is known that TM = R2 ×M . We provideM with the flat Riemannian metric. Namely, for all x ∈ M , the metric 〈 , 〉x on x is thestandard scalar product on R2 (which does not depend on x). Furthermore, we set thefollowing assumptions.

1) There is no closed geodesic of length ≤ 4.2) D is the disk of radius 1 centred at P1.3) d(P1, Pi) ≥ 2 for all i ∈ {n+ 1, . . . , m}.4) Let C1, . . . , Cq be the boundary components of M . Then d(P1, Cj) ≥ 2 for all

j ∈ {1, . . . , q}.Now, let f ∈ B1

m−n+1M . Let b = (b1, bn+1, . . . , bm) be a braid based at (P1,Pn+1, . . . , Pm) which represents f . For t ∈ [0, 1], we write

r(t) = inf{

12d(b1(t), bn+1(t)), . . . ,

12d(b1(t), bm(t)),

12d(b1(t), C1), . . . ,

12d(b1(t), Cq), 1

}.

Then r : [0, 1]→ R is a continuous map and r(t) > 0 for all t ∈ [0, 1]. Let

D0 = {(X1, X2) ∈ R2 ; X21 +X2

2 ≤ 1} .

Let H0 : D0 × [0, 1]→M be the map defined by

H0(X, t) = expb1(t)(r(t)X) for X ∈ D0 and for t ∈ [0, 1] .

Let F : D0 → D be the diffeomorphism defined by

F (X) = expP1X for X ∈ D0 .

LetH : D ∪ {Pn+1, . . . , Pm} × [0, 1] −→M

be the map defined by

H(x, t) = H0(F−1(x), t) for x ∈ D and for t ∈ [0, 1] ,H(Pi, t) = bi(t) for Pi ∈ {Pn+1, . . . , Pm} and for t ∈ [0, 1] .

23

The map H is a tunnel on M based at (D;Pn+1, . . . , Pm). We define ι0(f) to be theelement of Tm−nM represented by H.

One can easily verify that ι0 is well-defined, that ι0 is a morphism, and that κ◦ι0 = id.ut

Lemma 5.7. The morphism κ : Tm−nM → B1m−n+1M is surjective.

Remark. We do not know whether a similar result holds for non-orientable surfaces.

Proof. We choose an open disk K0 embedded in M \ D and which does not containany Pi for i = n + 1, . . . , m. The inclusion M \ K0 ⊆ M induces an epimorphism φ :B1m−n+1M \K0 → B1

m−n+1M . The following diagram commutes.

Tm−nM \K0κ−→ B1

m−n+1M \K0y yφTm−nM

κ−→ B1m−n+1M

By Lemma 5.6, κ : Tm−nM \ K0 → B1m−n+1M \ K0 is surjective. It follows that κ :

Tm−nM → B1m−n+1M is surjective, too. ut

From now on, we fix a (set) section ι0 : B1m−n+1M → Tm−nM of κ. Moreover, we

assume that ι0 is a morphism if M is either with non-empty boundary or a torus, and thatι0(1) = 1.

Theorem 5.1 is a direct consequence of the following lemma.

Lemma 5.8. Let n ≥ 2. Let g ∈ CBmM (BnD). There exist u ∈ B1m−n+1M and f ∈ BnD

such thatg = τ(ι0(u), f) .

The following lemmas 5.9 and 5.10 are preliminary results to the proof of Lemma 5.8.Recall that Σm denotes the group of permutations of {P1, . . . , Pm}, that Σn denotes

the group of permutations of {P1, . . . , Pn}, and that Σm−n denotes the group of permuta-tions of {Pn+1, . . . , Pm}. We write

BnmM = σ−1(Σm−n) .

Lemma 5.9. Let n ≥ 1. Let g ∈ CBnmM (PBnD). There exist u ∈ B1m−n+1M and

f ∈ PBnD such thatg = τ(ι0(u), f) .

24

Proof. We prove Lemma 5.9 by induction on n. Let n = 1. Then PB1D = {1}, thus

CB1mM

(PB1D) = B1mM .

On the other hand, if u ∈ B1mM , then

u = τ(ι0(u), P1) ,

where P1 denotes the constant path on P1.Let n > 1. Let g ∈ CBnmM (PBnD). We write M ′ = M \{P1, . . . , Pn−1, Pn+1, . . . , Pm},

and D′ = D \ {P1, . . . , Pn−1}. We consider the following commutative diagram.

1 −→ π1D′ −→ PBnD

ρ−→ PBn−1D −→ 1y y y1 −→ π1M

′ −→ BnmMρ−→ Bn−1

m−1M −→ 1

By Lemma 5.5, ρ(g) ∈ CBn−1m−1M

(PBn−1D). By induction, there exist u ∈ B1m−n+1M and

f1 ∈ PBn−1D such thatρ(g) = τ(ι0(u), f1) .

We choose f2 ∈ PBnD such that ρ(f2) = f1 and we write

g′ = g · τ(ι0(u), f2)−1 .

We have g′ ∈ π1M′ (since ρ(g′) = 1) and g′ ∈ CBnmM (PBnD), thus, by Lemma 5.5,

g′ ∈ Cπ1M ′(π1D′) .

If either m 6= n or M is not a disk, then M ′ is large and D′ is not a Mobius collar in M ′,thus, by Theorem 3.1,

Cπ1M ′(π1D′) = π1D

′ .

If m = n and M is a disk, then π1M′ = π1D

′, thus

Cπ1M ′(π1D′) = π1D

′ .

If follows thatg′ = f3 ∈ π1D

′ ⊆ PBnD .

So,g = f3 · τ(ι0(u), f2) = τ(ι0(u), f3f2) . ut

Lemma 5.10. Let n ≥ 2. Let g ∈ CBmM (BnD). Then σ(g) ∈ Σn ×Σm−n.

Proof. Let g ∈ CBmM (BnD). We suppose that σ(g)(Pn+1) = P1. Let f ∈π1(D \ {P2, . . . , Pn}, P1), f 6= 1. The group PBnD has finite index in BnD, thus

25

CBmM (BnD) = CBmM (PBnD). Since π1(D \ {P2, . . . , Pn}) ⊆ PBnD and since g ∈CBmM (PBnD), there exists an integer k > 0 such that

gfkg−1 ∈ PBnD .

We consider the following exact sequence.

1 −→ π1(M \ {P1, . . . , Pn, Pn+2, . . . , Pm}) −→ PBmMρ−→PBm−1M −→ 1

The morphism ρ sends PBnD isomorphically on PBnD. On the other hand, gfkg−1 6= 1(since f 6= 1 and BmM is torsion free) and ρ(gfkg−1) = 1 (see Figure 5.1). This is acontradiction.

This proves that σ(g) ∈ Σn ×Σm−n. ut

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n

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Figure 5.1

Proof of Lemma 5.8. Let g ∈ CBmM (BnD). By Lemma 5.10, σ(g) ∈ Σn × Σm−n. Wechoose f1 ∈ BnD such that σ(gf−1

1 ) ∈ Σm−n and we write g′ = gf−11 . Then g′ ∈ BnmM ,

g′ ∈ CBmM (BnD), and CBmM (BnD) = CBmM (PBnD), thus

g′ ∈ CBnmM (PBnD) .

By Lemma 5.9, there exist u ∈ B1m−n+1M and f2 ∈ PBnD such that

g′ = τ(ι0(u), f2) .

So,g = τ(ι0(u), f2) · f1 = τ(ι0(u), f2f1) . ut

Proof of Theorem 5.3. The proof of Theorem 5.3 is divided into 5 steps.

26

Step 1. Definition of δ.

We consider the natural morphism δ0 : BnmM → B1m−n+1M . Let g ∈ Cn,mM . By

Lemma 5.10, σ(g) ∈ Σn ×Σm−n. We choose f ∈ BnD such that σ(gf−1) ∈ Σm−n and weset

δ(g) = δ0(gf−1) .

We prove that the definition of δ(g) does not depend on the choice of f . Let f1, f2 ∈BnD be such that σ(gf−1

1 ) ∈ Σm−n and σ(gf−12 ) ∈ Σm−n. Then

δ0(gf−12 )−1δ0(gf−1

1 ) = δ0(f2g−1gf−1

1 ) = δ0(f2f−11 ) = 1 ,

thus δ0(gf−11 ) = δ0(gf−1

2 ).

Step 2. The map δ : Cn,mM → B1m−n+1M is a morphism.

Let g1, g2 ∈ Cn,mM . Let f1, f2 ∈ BnD be such that σ(g1f−11 ) ∈ Σm−n and σ(g2f

−12 ) ∈

Σm−n. By Corollary 5.2,Cn,mM = NBmM (BnD) ,

thus there exists f3 ∈ BnD such that g−12 f1g2 = f3. Moreover,

σ((g1g2)(f2f3)−1) = σ(g1f−11 g2f

−12 ) ∈ Σm−n .

So,

δ(g1)δ(g2) = δ0(g1f−11 )δ0(g2f

−12 ) = δ0(g1f

−11 g2f

−12 ) = δ0((g1g2)(f2f3)−1) = δ(g1g2) .

Step 3. Let h ∈ Tm−nM and let f ∈ BnD. Then

δ(τ(h, f)) = δ(τ(h, 1) · τ(1, f)) = δ(τ(h, 1)) · δ(f) = κ(h) .

Step 4. We have the following exact sequence.

1 −→ BnD −→ Cn,mMδ−→B1

m−n+1M −→ 1

Let u ∈ B1m−n+1M . Then

δ(τ(ι0(u), 1)) = κ(ι0(u)) = u .

This shows that δ is surjective.Let g ∈ Cn,mM . By Lemma 5.8, there exist u ∈ B1

m−n+1M and f ∈ BnD such thatg = τ(ι0(u), f). If g ∈ ker δ, then

1 = δ(g) = κ(ι0(u)) = u ,

27

thusg = τ(ι0(u), f) = τ(1, f) = f ∈ BnD .

Step 5. We have the following exact sequence.

1 −→ Z(BnD) −→ Zn,mMδ−→B1

m−n+1M −→ 1

By Step 4, it suffices to show that δ : Zn,mM → B1m−n+1M is surjective. Let u ∈

B1m−n+1M . Then τ(ι0(u), 1) ∈ Zn,mM and δ(τ(ι0(u), 1)) = u. ut

Proof of Theorem 5.4. The morphism ι : B1m−n+1M → Zn,mM is defined by

ι(u) = τ(ι0(u), 1) for u ∈ B1m−n+1M .

Clearly, δ ◦ ι = id. ut

6. Commensurator, normalizer, and

centralizer of BnN in BmM

Let M be a large surface, and let N be a subsurface of M such that N is neither adisk, nor a Mobius collar in M , and such that none of the connected components of M \Nis a disk. Let N1, . . . , Nr be the connected components of M \N . Let P1, . . . , Pn ∈ N ,and let Pn+1, . . . , Pm ∈M \N . For i = 1, . . . , r we write

Pi = {Pn+1, . . . , Pm} ∩Ni ,BniNi = BniNi(Pi) ,

where ni denotes the cardinality of Pi. If ni = 0, we make the convention that B0Ni = {1}.The goal of this section is to prove the following theorem.

Theorem 6.1.

CBmM (BnN) = BnN ×Bn1N1 × . . .×BnrNr .

Corollary 6.2.

CBmM (BnN) = NBmM (BnN) = BnN ×Bn1N1 × . . .×BnrNr ,ZBmM (BnN) = Z(BnN)×Bn1N1 × . . .×BnrNr .

The remains of this section are divided into two subsections. In Subsection 6.1 westudy an action of π1N on some groupoid Π1(M \ {P0}). In Subsection 6.2 we apply theresults of Subsection 6.1 to prove Theorem 6.1.

28

6.1. Action of π1N on Π1(M \ {P0}).

Throughout this subsection, we fix a point P0 ∈ N and a point Pi ∈ Ni for alli = 1, . . . , r. Moreover, we do not assume that none of the connected components ofM \N is a disk.

The fundamental groupoid of M \ {P0} based at {P1, . . . , Pr} is the groupoid Π1(M \{P0}) defined by the following data.

1) The set of objects of Π1(M \ {P0}) is {P1, . . . , Pr}.2) Let Pi, Pj ∈ {P1, . . . , Pr}. The set of morphisms from Pi to Pj is the set Π1(M \

{P0})[Pi, Pj] of homotopy classes of paths in M \ {P0} from Pi to Pj .Let Pi, Pj, Pk ∈ {P1, . . . , Pr}. For convenience, we assume that the composition map

goes from Π1(M \ {P0})[Pi, Pj ]×Π1(M \ {P0})[Pj, Pk] to Π1(M \ {P0})[Pi, Pk]. Note that

Π1(M \ {P0})[Pi, Pi] = π1(M \ {P0}, Pi) .

Moreover, if x ∈ Π1(M \ {P0})[Pi, Pj], then the map

θx : π1(M \ {P0}, Pi) −→ Π1(M \ {P0})[Pi, Pj]g 7−→ gx

is a bijection.Let Pi, Pj ∈ {P1, . . . , Pr}. An interbraid on M based at (P0, [Pi, Pj]) is a pair b =

(b0, b1) of paths, bk : [0, 1]→M , such that1) b0(0) = b0(1) = P0, b1(0) = Pi, and b1(1) = Pj ,2) b0(t) 6= b1(t) for t ∈ [0, 1].

There is a natural notion of homotopy of interbraids. The interbraid groupoid on M basedat (P0, {P1, . . . , Pr}) is the groupoid IB2M = IB2M(P0, {P1, . . . , Pr}) defined by thefollowing data.

1) The set of objects of IB2M is {P1, . . . , Pr}.2) Let Pi, Pj ∈ {P1, . . . , Pr}. The set of morphisms from Pi to Pj is the set

IB2M [Pi, Pj] of homotopy classes of interbraids on M based at (P0, [Pi, Pj]).Let Pi, Pj, Pk ∈ {P1, . . . , Pr}. For convenience, we assume that the composition map

goes from IB2M [Pi, Pj]× IB2M [Pj , Pk] to IB2M [Pi, Pk]. Note that

IB2M [Pi, Pi] = PB2M(P0, Pi) .

Moreover, if X ∈ IB2M [Pi, Pj], then the map

ΘX : PB2M(P0, Pi) −→ IB2M [Pi, Pj]g 7−→ gX

is a bijection.Let Pi, Pj ∈ {P1, . . . , Pr}. We consider the natural maps

α : IB2M [Pi, Pj ] −→ π1(M,P0) ,β : Π1(M \ {P0})[Pi, Pj] −→ IB2M [Pi, Pj] .

29

Let x ∈ Π1(M \ {P0})[Pi, Pj], and let X = β(x). Then the following diagram commutes.

1 −→ π1(M \ {P0}, Pi) −→ PB2M(P0, Pi)ρ−→ π1(M,P0) −→ 1yθx yΘX

yid

Π1(M \ {P0})[Pi, Pj]β−→ IB2M [Pi, Pj]

α−→ π1(M,P0)

Thus, α is surjective, β is injective, and

α−1(1) = β(Π1(M \ {P0})[Pi, Pj]) .

So, we can assume that

Π1(M \ {P0})[Pi, Pj] = β(Π1(M \ {P0})[Pi, Pj]) ⊆ IB2M [Pi, Pj] .

The inclusion N ⊆ M induces a morphism ψk : π1(N,P0) → PB2M(P0, Pk) for allk = 1, . . . , r. We define an action of π1(N,P0) on IB2M [Pi, Pj] as follows. Let u ∈π1(N,P0) and let X ∈ IB2M [Pi, Pj ]. Then

u(X) = ψi(u) ·X · ψj(u)−1 .

Let x ∈ Π1(M \ {P0})[Pi, Pj ] and let u ∈ π1(N,P0). Then α(u(x)) = 1, thus u(x) ∈Π1(M \ {P0})[Pi, Pj]. So, the action of π1(N,P0) on IB2M [Pi, Pj ] induces an action ofπ1(N,P0) on Π1(M \ {P0})[Pi, Pj ].

We denote by SN [Pi, Pj] the set of x ∈ Π1(M \ {P0})[Pi, Pj] such that, for all u ∈π1(N,P0) there exists an integer k > 0 such that uk(x) = x. The main result of Subsection6.1 is the following proposition.

Proposition 6.3. Let i, j ∈ {1, . . . , r}.

SN [Pi, Pj] =

{π1(Ni, Pi) = π1(Nj , Pj) if i = j ,

∅ if i 6= j .

The lemmas 6.4 to 6.7 are preliminary results to the proof of Proposition 6.3.From now on and till the end of the proof of Lemma 6.7, we set the following assump-

tions (see Figure 6.1).1) N is a sphere with q + 1 holes (q ≥ 1). We denote by C0, C1, . . . , Cq the boundary

components of N .2) M \N has two connected components, N1 and N2.3) N ∩N1 = C1 ∪ . . . ∪ Cq, and N ∩N2 = C0.

30

0

N N N1 2

C

C

C

C

1

2

q

Figure 6.1

We choose a point P ′0 ∈ N different from P0. We choose a point Qi ∈ Ci for all i =0, 1, . . . , q. According to Figure 6.2,

1) we choose a path γsi : [0, 1]→ N \ {P0} from P ′0 to Qi for all i = 0, 1, . . . , q,2) we choose a path γti : [0, 1]→ N1 from P1 to Qi for all i = 1, . . . , q,3) we choose a path γt0 : [0, 1]→ N2 from P2 to Q0.

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γ

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qt

qs

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0

Figure 6.2

We writeγi = γsi (γ

ti )−1 for i = 0, 1, . . . , q ,

βi = γ−11 γi ∈ π1(M \ {P0}, P1) for i = 1, . . . , q ,

T = γ−11 γ0 ∈ Π1(M \ {P0})[P1, P2] .

Note that the path T induces a morphism

π1(N2, P2) −→ π1(M \ {P0}, P1)g 7−→ TgT−1

31

The following lemma is a consequence of Van Kampen’s theorem.

Lemma 6.4. Let F be the subgroup of π1(M \ {P0}, P1) generated by β2, . . . , βq.

π1(M \ {P0}, P1) = π1(N1, P1) ∗ (T · π1(N2, P2) · T−1) ∗ F .

All these groups are free and {β2, . . . , βq} is a basis for F . ut

According to Figure 6.3,1) we choose a simple loop αi : [0, 1]→ Ci based at Qi for all i = 0, 1, . . . , q,2) we choose a path δi : [0, 1]→ N from P0 to Qi for all i = 0, 1, . . . , q.

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2

1

0

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2

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P’0

Figure 6.3

We writehi = γtiαi(γ

ti )−1 ∈ π1(N1, P1) for i = 1, . . . , q ,

h0 = γt0α0(γt0)−1 ∈ π1(N2, P2) ,

ui = δiαiδ−1i ∈ π1(N,P0) for i = 0, 1, . . . , q .

According to Figure 6.4, we choose a loop µ : [0, 1]→ N \{P0} based at P ′0 turning aroundP0.

32

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µ

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P0

Figure 6.4

We writehc = γ−1

1 µγ1 ∈ π1(M \ {P0}, P1) .

One can easily verify that

hc = Th−10 T−1 · h−1

1 · β2h−12 β−1

2 · . . . · βqh−1q β−1

q .

Lemma 6.5. i) u0(g) = g for all g ∈ π1(N1, P1).ii) u0(g) = g for all g ∈ π1(N2, P2).iii) u0(βi) = βi for all βi ∈ {β2, . . . , βq}.iv) u0(T ) = h−1

c T .

Proof. i) We choose a loop ζ : [0, 1]→ N1 based at P1 which represents g. Then the imageof ζ and the image of u0 are disjoint (see Figure 6.5), thus u0(g) = g.

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ii) We choose a loop ζ : [0, 1] → N2 based at P2 which represents g. The image of ζand the image of u0 are disjoint, thus u0(g) = g.

iii) The image of βi and the image of u0 are disjoint, thus u0(βi) = βi.

33

iv) In Figure 6.6, the interbraid drawn in (a) is homotopic to the interbraid drawnin (b), and the interbraid drawn in (b) is homotopic to the interbraid drawn in (c). Theinterbraid drawn in (a) represents u0(T ), and the interbraid drawn in (c) represents

γ−11 µ−1γ0 .

It follows that

u0(T ) = γ−11 µ−1γ0 = γ−1

1 µ−1γ1 · γ−11 γ0 = h−1

c T . ut

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34

Lemma 6.6. Let k ∈ {2, . . . , q}.i) uk(g) = g for all g ∈ π1(N1, P1).ii) uk(g) = g for all g ∈ π1(N2, P2).iii) uk(T ) = T .iv) uk(βi) = βi for all i ∈ {2, . . . , k − 1}.v) uk(βk) = βkh

−1k β−1

k h−1c βkhk.

vi) uk(hc) = βkh−1k β−1

k hcβkhkβ−1k .

Proof. The statements i) to iv) can be proved with the same arguments as those given inthe proofs of the statements i) to iii) of Lemma 6.5.

v) In Figure 6.7, the braid drawn in (a) is homotopic to the braid drawn in (b), andthe braid drawn in (b) is homotopic to the braid drawn in (c). The braid drawn in (a)represents uk(βk), and the braid drawn in (c) represents

γ−11 γskα

−1k (γsk)−1µ−1γskαk(γtk)−1 .

It follows that

uk(βk) = γ−11 γskα

−1k (γsk)−1µ−1γskαk(γtk)−1

= γ−11 γsk(γtk)−1 · γtkα−1

k (γtk)−1 · γtk(γsk)−1γ1 · γ−11 µ−1γ1 · γ−1

1 γsk(γtk)−1 · γtkαk(γtk)−1

= βkh−1k β−1

k h−1c βkhk

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35

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Figure 6.7.c

vi) In Figure 6.8, the braid drawn in (a) is homotopic to the braid drawn in (b), thebraid drawn in (b) is homotopic to the braid drawn in (c), and the braid drawn in (c) ishomotopic to the braid drawn in (d). The braid drawn in (a) represents uk(hc), and thebraid drawn in (d) represents

γ−11 γskα

−1k (γsk)−1µγskαk(γsk)−1γ1 .

It follows that

uk(hc) = γ−11 γskα

−1k (γsk)−1µγskαk(γsk)−1γ1

= γ−11 γsk(γtk)−1 · γtkα−1

k (γtk)−1 · γtk(γsk)−1γ1 · γ−11 µγ1 · γ−1

1 γsk(γtk)−1·γtkαk(γtk)−1 · γtk(γsk)−1γ1

= βkh−1k β−1

k hcβkhkβ−1k ut

36

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Lemma 6.7. SN [P1, P1] = π1(N1, P1) and SN [P1, P2] = ∅.

Proof. The proof of Lemma 6.7 is divided into 5 steps.

Step 1. π1(N1, P1) ⊆ SN [P1, P1].

Let g ∈ π1(N1, P1) and let u ∈ π1(N,P0). Let ζ : [0, 1] → N1 be a loop based at P1

which represents g, and let ξ : [0, 1] → N be a loop based at P0 which represents u. Theimage of ζ and the image of ξ are disjoint, thus u(g) = g.

Step 2. SN [P1, P1] ⊆ π1(N1, P1) ∗ F .

38

Leth′c = βqhqβ

−1q · . . . · β2h2β

−12 · h1 .

Then

hc = Th−10 T−1 · (h′c)−1 ,

h′c ∈ π1(N1, P1) ∗ F .

Let g ∈ π1(M \ {P0}, P1). By Lemma 6.4, g can be (uniquely) written

g = x0Ty1T−1x1 . . . TylT

−1xl ,

wherexi ∈ π1(N1, P1) ∗ F for i = 0, 1, . . . , l ,xi 6= 1 for i = 1, . . . , l− 1 ,yi ∈ π1(N2, P2) \ {1} for i = 1, . . . , l .

We suppose that l ≥ 1. By Lemma 6.5,

u0(g) = x0h−1c Ty1T

−1hcx1 . . . h−1c TylT

−1hcxl

= x0h′c · T · h0y1h

−10 · T−1 · (h′c)−1x1h

′c · . . . · T · h0ylh

−10 · T−1 · (h′c)−1xl

It follows that, for an integer k > 0,

uk0(g) = x0(h′c)k · T · hk0y1h

−k0 · T−1 · (h′c)−kx1(h′c)

k · . . . · T · hk0ylh−k0 · T−1 · (h′c)−kxl ,

thus uk0(g) 6= g.So, if g ∈ SN [P1, P1], then there exists an integer k > 0 such that uk0(g) = g, thus

l = 0, therefore g ∈ π1(N1, P1) ∗ F .

For j = 2, . . . , q, we denote by F (β2, . . . , βj) the subgroup of F generated by {β2, . . . ,βj}.

Step 3. SN [P1, P1] ⊆ π1(N1, P1) ∗ F (β2, . . . , βq−1).

Leth′ = βq−1hq−1β

−1q−1 · . . . · β2h2β

−12 · h1 · Th0T

−1 .

Then

hc = (h′)−1βqh−1q β−1

q ,

h′ ∈ π1(N1, P1) ∗ (T · π1(N2, P2) · T−1) ∗ F (β2, . . . , βq−1) .

Let g ∈ π1(M \ {P0}, P1). By Lemma 6.4, g can be (uniquely) written

g = x0βε1q x1 . . . β

εlq xl ,

39

where

xi ∈ π1(N1, P1) ∗ (T · π1(N2, P2) · T−1) ∗ F (β2, . . . , βq−1) for i = 0, 1, . . . , l ,εi ∈ {±1} for i = 1, . . . , l ,xi 6= 1 if εi+1 = −εi for i = 1, . . . , l− 1 .

We call this expression a relative reduced expression of g with respect to βq of lengthl = lq(g).

We suppose that l ≥ 1. Let k > 0 be an integer. By Lemma 6.6,

uq(βq) = βqh−1q β−1

q h−1c βqhq = h′βqhq ,

uq(xi) = xi for i = 0, 1, . . . , l .

If εi = εi+1 = 1, then

ukq (βqxiβq) = (h′)k · βq · hkqxi(h′)k · βq · hkq .

If εi = 1 and εi+1 = −1, then

ukq (βqxiβ−1q ) = (h′)k · βq · hkqxih−kq · β−1

q · (h′)−k ,

and hkqxih−kq 6= 1 (since xi 6= 1). If εi = −1 and εi+1 = 1, then,

ukq (β−1q xiβq) = h−kq · β−1

q · (h′)−kxi(h′)k · βq · hkq ,

and (h′)−kxi(h′)k 6= 1 (since xi 6= 1). If εi = εi+1 = −1, then

ukq (β−1q xiβ

−1q ) = h−kq · β−1

q · (h′)−kxih−kq · β−1q · (h′)−k .

So, ukq (g) has a relative reduced expression with respect to βq of length l, and this expressionbegins with either x0(h′)k (if ε1 = 1) or x0h

−kq (if ε1 = −1). In particular, ukq (g) 6= g.

So, if g ∈ SN [P1, P1], then there exists an integer k > 0 such that ukq (g) = g, thuslq(g) = 0, therefore

g ∈ π1(N1, P1) ∗ (T · π1(N2, P2) · T−1) ∗ F (β2, . . . , βq−1) .

By Step 2, it follows that

g ∈ π1(N1, P1) ∗ F (β2, . . . , βq−1) .

Step 4. SN [P1, P1] ⊆ π1(N1, P1).

By Step 3,SN [P1, P1] ⊆ π1(N1, P1) ∗ F (β2, . . . , βq−1) .

40

Let j ∈ {2, . . . , q − 1}. We suppose that SN [P1, P1] ⊆ π1(N1, P1) ∗ F (β2, . . . , βj) and weprove that SN [P1, P1] ⊆ π1(N1, P1) ∗ F (β2, . . . , βj−1).

Let R be the set of g ∈ π1(M \ {P0}, P1) which can be (uniquely) written

g = x0βε1j x1 . . . β

εlj xl ,

where

either xi ∈ π1(N1, P1) ∗ F (β2, . . . , βj−1) or xi ∈ {hc, h−1c } for i = 0, 1, . . . , l ,

xi 6= 1 if εi+1 = −εi for i = 1, . . . , l− 1 ,

x0, xl 6∈ {hc, h−1c } ,

εi = −1 and εi+1 = 1 if xi ∈ {hc, h−1c } for i = 1, . . . , l− 1 .

We write l = lR(g).In order to be able to choose j ∈ {2, . . . , q−1}, we first have to assume that q ≥ 3. In

particular, neither N1, nor N∪N2 is a disk, thus hi 6= 1 for all i = 1, . . . , q. The uniquenessof the expression of g comes from the fact that hc can be written

hc = Th−10 T−1 · h−1

1 · β2h−12 β−1

2 · . . . · βjh−1j β−1

j · βj+1h−1j+1β

−1j+1 · . . . · βqh−1

q β−1q .

This kind of expression would not be necessarily unique if j = q.We suppose that l ≥ 1. If εi = εi+1 = 1, then, by Lemma 6.6,

uj(βjxiβj) = βj · h−1j · β−1

j · h−1c · βj · hjxi · βj · h−1

j · β−1j · h−1

c · βj · hj .

If εi = 1 and εi+1 = −1, then, by Lemma 6.6,

uj(βjxiβ−1j ) = βj · h−1

j · β−1j · h−1

c · βj · hjxih−1j · β−1

j · hc · βj · hj · β−1j ,

and hjxih−1j 6= 1 (since xi 6= 1). If εi = −1, εi+1 = 1, and xi 6∈ {hc, h−1

c }, then, by Lemma6.6,

uj(β−1j xiβj) = h−1

j · β−1j · hc · βj · hj · β−1

j · xi · βj · h−1j · β−1

j · h−1c · βj · hj .

If εi = εi+1 = −1, then, by Lemma 6.6,

uj(β−1j xiβ

−1j ) = h−1

j · β−1j · hc · βj · hj · β−1

j · xih−1j · β−1

j · hc · βj · hj · β−1j .

If εi = −1, εi+1 = 1 and xi = hεc (where ε ∈ {±1}), then, by Lemma 6.6,

uj(β−1j hεcβj) = h−1

j β−1j hcβjhjβ

−1j · βjh−1

j β−1j hεcβjhjβ

−1j · βjh−1

j β−1j h−1

c βjhj

= h−1j · β−1

j · hεc · βj · hj

It follows that uj(g) ∈ R, that lR(uj(g)) ≥ l, and that lR(uj(g)) > l if none of the xi isincluded in {hc, h−1

c } for i = 1, . . . , l− 1. This shows that ukj (g) 6= g if k > 0 is an integer,if g ∈ π1(N1, P1) ∗ F (β2, . . . , βj), and if lR(g) ≥ 1.

41

Let g ∈ SN [P1, P1]. By hypothesis, g ∈ π1(N1, P1)∗F (β2, . . . , βj). There exists an in-teger k > 0 such that ukj (g) = g, thus lR(g) = 0, therefore g ∈ π1(N1, P1)∗F (β2, . . . , βj−1).

Step 5. SN [P1, P2] = ∅.

Let g ∈ Π1(M \ {P0})[P1, P2]. By Lemma 6.4, g can be (uniquely) written

g = x0Ty1T−1x1 . . . TylT

−1xlT ,

wherexi ∈ π1(N1, P1) ∗ F for i = 0, 1, . . . , l ,xi 6= 1 for i = 1, . . . , l− 1 ,yi ∈ π1(N2, P2) \ {1} for i = 1, . . . , l .

By Lemma 6.5,

u0(g) = x0h−1c Ty1T

−1hcx1 . . . h−1c TylT

−1hcxlh−1c T

= x0h′c · T · h0y1h

−10 · T−1 · (h′c)−1x1h

′c · . . . · T · h0ylh

−10 · T−1 · (h′c)−1xlh

′c · T · h0

It follows that, for an integer k > 0,

uk0(g) =x0(h′c)k · T · hk0y1h

−k0 · T−1 · (h′c)−kx1(h′c)

k · . . . · T · hk0ylh−k0 · T−1·(h′c)

−kxl(h′c)k · T · hk0 ,

thus uk0(g) 6= g. ut

Now, the special assumptions on N that we made just before Proposition 6.3 aredropped.

Proof of Proposition 6.3. We prove that SN [P1, P1] = π1(N1, P1) and that SN [P1, P2] = ∅.The same argument works for any Pi and Pj .

Let g ∈ π1(N1, P1) and let u ∈ π1(N,P0). Let ζ : [0, 1] → N1 be a loop based atP1 which represents g, and let ξ : [0, 1] → N be a loop based at P0 which representsu. The image of ζ and the image of ξ are disjoint, thus u(g) = g. This shows thatπ1(N1, P1) ⊆ SN [P1, P1].

Now, let C1, . . . , Cq be the connected components of N ∩N1. We choose a subsurfaceN ′ ⊆ N (see Figure 6.9) such that

1) N ′ is a sphere with q + 1 holes,2) M \N ′ has two connected components, N1 and

N ′2 = N \N ′ ∪N2 ∪ . . . ∪Nr ,

3) N ′ ∩N1 = C1 ∪ . . . ∪ Cq,4) N ′ ∩N ′2 has a unique connected component that we denote by C0,5) P0 ∈ N ′.

42

Moreover, in the case where r = 1, we pick some point P2 ∈ N ′2.

��������

��������

N’

N N

P P

1

10

C

C

C

C

1

2

q

0

Figure 6.9

Let SN ′ [P1, P1] be the set of g ∈ π1(M \ {P0}, P1) such that, for all u ∈ π1(N ′, P0),there exists an integer k > 0 such that uk(g) = g. We have SN [P1, P1] ⊆ SN ′ [P1, P1] (sinceN ⊇ N ′), and, by Lemma 6.7, SN ′ [P1, P1] = π1(N1, P1), thus SN [P1, P1] ⊆ π1(N1, P1). Itis clear by disjointness that π1(N1, P1) ⊆ SN [P1, P1], so π1(N1, P1) = SN [P1, P1].

Now, we assume that r ≥ 2. Let SN ′ [P1, P2] be the set of g ∈ Π1(M \ {P0})[P1, P2]such that, for all u ∈ π1(N ′, P0), there exists an integer k > 0 such that uk(g) = g. Wehave SN [P1, P2] ⊆ SN ′ [P1, P2] (since N ⊇ N ′), and, by Lemma 6.7, SN ′ [P1, P2] = ∅, thusSN [P1, P2] = ∅. ut

6.2. Proof of Theorem 6.1

The lemmas 6.8 to 6.12 are preliminary results to the proof of Theorem 6.1.

Lemma 6.8. Let m = 2 and let n = 1. Let i ∈ {1, . . . , r} be such that P2 ∈ Ni. Then

CPB2M (PB1N) = CPB2M (π1N) = π1(N,P1)× π1(Ni, P2) .

Proof. We assume that i = 1 (i.e. P2 ∈ N1). The inclusion

π1(N,P1)× π1(N1, P2) ⊆ CPB2M (π1N)

is obvious.Let g ∈ CPB2M (π1N). We consider the following exact sequence.

1 −→ π1(M \ {P1}) −→ PB2Mρ−→π1M −→ 1

43

The morphism ρ sends π1(N,P1) isomorphically on π1(N,P1). By Lemma 5.5, ρ(g) ∈Cπ1M (π1N). By Theorem 3.1, ρ(g) = f ∈ π1(N,P1). We write g′ = f−1g. We haveg′ ∈ π1(M \ {P1}, P2) (since ρ(g′) = 1) and g′ ∈ CPB2M (π1N).

Let u ∈ π1(N,P1). Since g′ ∈ CPB2M (π1N), there exists an integer k > 0 such that

g′uk(g′)−1 ∈ π1(N,P1) .

The morphism ρ sends π1(N,P1) isomorphically on π1(N,P1), thus

g′uk(g′)−1 = ρ(g′uk(g′)−1) = ρ(g′)ρ(uk)ρ(g′)−1 = uk ,

thereforeukg′u−k = g′ .

We write Q1 = P2 and we choose a point Qi ∈ Ni for all i = 2, . . . , r. Let Π1(M \{P1})be the fundamental groupoid on M \ {P1} based at {Q1, Q2, . . . , Qr}. By the aboveconsiderations, g′ ∈ SN [Q1, Q1], thus, by Proposition 6.3, g′ ∈ π1(N1, P2). So,

g = fg′ ∈ π1(N,P1)× π1(N1, P2) . ut

Lemma 6.9.

CPBmM (PBnN) = PBnN × PBn1N1 × . . .× PBnrNr .

Proof. The proof of Lemma 6.9 is divided into 2 steps.

Step 1. Let n = 1. We prove by induction on m that

CPBmM (PB1N) = CPBmM (π1N) = π1N × PBn1N1 × . . .× PBnrNr .

The case m = 1 is proved in Theorem 3.1, and the case m = 2 is proved in Lemma6.8. Let m ≥ 3. The inclusion

π1N × PBn1N1 × . . .× PBnrNr ⊆ CPBmM (π1N)

is obvious.Let g ∈ CPBmM (π1N). We consider the following exact sequence.

1 −→ π1(M \ {P1, P2, . . . , Pm−1}) −→ PBmMρ−→PBm−1M −→ 1

The morphism ρ sends π1(N,P1) isomorphically on π1(N,P1). By Lemma 5.5, ρ(g) ∈CPBm−1M (π1N). We assume that Pm ∈ N1. By induction,

CPBm−1M (π1N) = π1N × PBn1−1N1 × PBn2N2 × . . .× PBnrNr .

44

Thus we can choose f ∈ π1(N,P1), h′1 ∈ PBn1−1N1, and hi ∈ PBniNi for all i = 2, . . . , rsuch that

ρ(g) = fh′1h2 . . . hr .

The morphism ρ sends PBniNi isomorphically on PBniNi for all i = 2, . . . , r, and sendsPBn1N1 surjectively on PBn1−1N1. We choose h1 ∈ PBn1N1 such that ρ(h1) = h′1 andwe write

g′ = gh−1r . . . h−1

2 h−11 f−1 .

We have g′ ∈ π1(M \ {P1, P2, . . . , Pm−1}, Pm) (since ρ(g′) = 1) and g′ ∈ CPBmM (π1N).We have the inclusions

π1(M \ {P1, P2, . . . , Pm−1}) ⊆ PB2M \ {P2, . . . , Pm−1} ,π1N ⊆ PB2M \ {P2, . . . , Pm−1} ,

where PB2M \{P2, . . . , Pm−1} denotes the pure braid group on M \{P2, . . . , Pm−1} basedat (P1, Pm). So,

g′ ∈ CPB2M\{P2,...,Pm−1}(π1N) ,

thus, by Lemma 6.6,g′ ∈ π1(N,P1)× π1(N1 \ P ′1, Pm) ,

where P ′1 = P1 \ {Pm}. Let f ∈ π1(N,P1) and let h1 ∈ π1(N1 \ P ′1, Pm) be such thatg′ = f h1. Then

1 = ρ(g′) = ρ(f)ρ(h1) = f ,

thus g′ = h1 ∈ π1(N1 \ P ′1, Pm). So,

g = g′ · fh1h2 . . . hr = f(g′h1)h2 . . . hr ∈ π1N × PBn1N1 × PBn2N2 × . . .× PBnrNr .

Step 2. We prove by induction on n that

CPBmM (PBnN) = PBnN × PBn1N1 × . . .× PBnrNr .

The case n = 1 is proved in Step 1. Let n > 1. The inclusion

PBnN × PBn1N1 × . . .× PBnrNr ⊆ CPBmM (PBnN)

is obvious.Let g ∈ CPBmM (PBnN). Let N ′ = N \ {P1, . . . , Pn−1} and let M ′ = M \ {P1, . . . ,

Pn−1, Pn+1, . . . , Pm}. We consider the following commutative diagram.

1 −→ π1N′ −→ PBnN

ρ−→ PBn−1N −→ 1y y y1 −→ π1M

′ −→ PBmMρ−→ PBm−1M −→ 1

45

By Lemma 5.5, ρ(g) ∈ CPBm−1M (PBn−1N). By induction,

CPBm−1M (PBn−1N) = PBn−1N × PBn1N1 × . . .× PBnrNr .

Thus we can choose f ′ ∈ PBn−1N and hi ∈ PBniNi for all i = 1, . . . , r such that

ρ(g) = f ′h1 . . . hr .

The morphism ρ sends PBniNi isomorphically on PBniNi for all i = 1, . . . , r, and sendsPBnN surjectively on PBn−1N . We choose f ∈ PBnN such that ρ(f) = f ′ and we write

g′ = gh−1r . . . h−1

1 f−1 .

We have g′ ∈ π1(M ′, Pn) (since ρ(g′) = 1) and g′ ∈ CPBmM (PBnN), thus, by Lemma 5.5,g′ ∈ Cπ1M ′(π1N

′). By Theorem 3.1, g′ ∈ π1(N ′, Pn) ⊆ PBnN . So,

g = (g′f)h1 . . . hr ∈ PBnN × PBn1N1 × . . .× PBnrNr . ut

Lemma 6.10. Let m = n. Then

CBnM (BnN) = BnN .

Proof. The inclusionBnN ⊆ CBnM (BnN)

is obvious.Let g ∈ CBnM (BnN). We choose f ∈ BnN such that σ(f) = σ(g) and we write

g′ = gf−1. We have g′ ∈ PBnM and g′ ∈ CBnM (BnN) = CBnM (PBnN), thus g′ ∈CPBnM (PBnN). By Lemma 6.9, g′ ∈ PBnN . So,

g = g′f ∈ BnN . ut

Recall that Σm denotes the group of permutations of {P1, . . . , Pm}, that Σn denotesthe group of permutations of {P1, . . . , Pn}, and that Σm−n denotes the group of permuta-tions of {Pn+1, . . . , Pm}. The following lemma can be proved with the same arguments asthose given in the proof of Lemma 5.10. Note that, since π1(N,P1) 6= {1}, we do not needto assume that n ≥ 2 in Lemma 6.11.

Lemma 6.11. Let g ∈ CBmM (BnN). Then σ(g) ∈ Σn ×Σm−n. ut

Let Σni denote the group of permutations of Pi for i = 1, . . . , r.

46

Lemma 6.12. Let g ∈ CBmM (BnN). Then

σ(g) ∈ Σn ×Σn1 × . . .× Σnr .

Proof. Let g ∈ CBmM (BnN). By Lemma 6.11, g ∈ σ−1(Σn × Σm−n). We consider thefollowing exact sequence.

1 −→ Bm−nM \ {P1, . . . , Pn} −→ σ−1(Σn ×Σm−n)ρ−→BnM −→ 1

The morphism ρ sends BnN isomorphically on BnN . By Lemma 5.5, ρ(g) ∈ CBnM (BnN).By Lemma 6.10, ρ(g) = f ∈ BnN . We write g′ = gf−1. We have g′ ∈ Bm−nM \{P1, . . . , Pn} (since ρ(g′) = 1) and g′ ∈ CBmM (BnN).

Let h ∈ BnN . Since g′ ∈ CBmM (BnN), there exists an integer k > 0 such that

g′hk(g′)−1 ∈ BnN .

Since ρ is an isomorphism on BnN ,

g′hk(g′)−1 = ρ(g′hk(g′)−1) = ρ(g′)ρ(hk)ρ(g′)−1 = hk ,

thereforehkg′h−k = g′ .

We suppose that Pn+1 ∈ N1, that Pn+2 ∈ N2, and that σ(g)(Pn+1) = Pn+2. Wealso have σ(g′)(Pn+1) = Pn+2. We write Q1 = Pn+1 and Q2 = Pn+2. We choose a pointQi ∈ Ni for all i = 3, . . . , r. Let Π1(M \ {P1}) be the fundamental groupoid on M \ {P1}based at {Q1, . . . , Qr}. Let b′ = (b′n+1, . . . , b

′m) be a braid on M \ {P1, . . . , Pn} based at

(Pn+1, . . . , Pm) which represents g′. Let x ∈ Π1(M \{P1})[Q1, Q2] be represented by b′n+1.By the above considerations, x ∈ SN [Q1, Q2]. This contradicts Proposition 6.3.

So,σ(g) ∈ Σn ×Σn1 × . . .×Σnr . ut

Proof of Theorem 6.1. The inclusion

BnN ×Bn1N1 × . . .×BnrNr ⊆ CBmM (BnN)

is obvious.Let g ∈ CBmM (BnN). By Lemma 6.12,

σ(g) ∈ Σn ×Σn1 × . . .×Σnr .

Thus we can choose f ∈ BnN and hi ∈ BniNi for all i = 1, . . . , r such that

σ(g) = σ(f)σ(h1) . . . σ(hr) .

47

We writeg′ = gh−1

r . . . h−11 f−1 .

We have g′ ∈ PBmM and g′ ∈ CBmM (BnN) = CBmM (PBnN), thus g′ ∈CPBmM (PBnN). By Lemma 6.9, there exist f ′ ∈ PBnN and h′i ∈ PBniNi for alli = 1, . . . , r such that

g′ = f ′h′1 . . . h′r .

So,g = f ′h′1 . . . h

′r · fh1 . . . hr = (f ′f)(h′1h1) . . . (h′rhr)

∈ BnN ×Bn1N1 × . . .×BnrNr . ut

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Authors addresses

Luis ParisUniversite de BourgogneLaboratoire de TopologieUMR 5584 du CNRSBP 40021011 Dijon CedexFRANCE

lparis@satie.u-bourgogne.fr

Dale RolfsenDepartment of MathematicsUniversity of British ColumbiaVancouverBritish Columbia V6T 1Z2CANADA

rolfsen@math.ubc.ca

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